The Power Rule: Statement and Proof
The power rule is the first and most-used derivative fact. These notes prove it honestly from the limit definition, so the rule is a theorem you own rather than a pattern you memorised.
Calculus · Differentiation
1. The statement
proved here for positive integers n; it extends to all real exponents via logarithmic differentiation.
2. Proof from the limit definition
Start from the definition of the derivative and expand (x+h)ⁿ with the binomial theorem:
Subtract xⁿ and divide by h:
As h → 0 the underbraced tail vanishes term by term, leaving exactly nxⁿ⁻¹. The reasoning to remember: only the linear-in-h part of the expansion survives the limit — the derivative is precisely the coefficient of h.
3. Extension by linearity
Because differentiation is linear, the rule handles every polynomial at once:
Evaluate any single-term case numerically with the power-rule calculator, which applies a·n·xⁿ⁻¹ with the substitution shown.
Key takeaway
The power rule is the binomial theorem with the limit filtering out everything but the first-order term — a pattern (linearise, keep the h-coefficient) that recurs across all of differential calculus.
4. Extending the proof beyond positive integers
Negative exponents. For y = x^{−m} (m a positive integer, x ≠ 0), run the limit definition directly:
using the positive-integer result in the numerator. The rule's form nxⁿ⁻¹ survives with n = −m ∎.
Rational exponents. Let y = x^{p/q} (x > 0), so y^q = x^p. Differentiate both sides implicitly: q y^{q−1}y′ = p x^{p−1}, hence
All real exponents. Write xⁿ = e^{n\ln x} and use the chain rule: d/dx e^{n ln x} = e^{n ln x}·(n/x) = n xⁿ⁻¹ — the definitive proof, covering n = √2 or π, at the price of assuming the calculus of e^x and ln x (built independently, so no circularity).
5. More worked examples
Example 2 — a tangent line. f(x) = x³ at x = 2: f′(x) = 3x², slope f′(2) = 12, point (2, 8): tangent y = 8 + 12(x − 2) = 12x − 16. The rule's daily job description.
Example 3 — roots and reciprocals without new rules. d/dx √x = d/dx x^{1/2} = ½x^{−1/2} = 1/(2√x); d/dx (1/x²) = −2x^{−3}; d/dx x^{5/3} = (5/3)x^{2/3}. Rewriting into power form first is the whole technique — students who differentiate 1/x² "by quotient rule" are doing triple the work for the same answer.
Example 4 — physics in one line. Position s(t) = 4.9t² (free fall, metres): velocity s′ = 9.8t, acceleration s″ = 9.8 m/s² — the constant g recovered by two applications of the rule. Kinematics is the power rule applied to polynomials in t.
6. Common misconceptions
"d/dx xˣ = x·xˣ⁻¹." The rule requires a constant exponent. For xˣ, take logs: y = xˣ → ln y = x ln x → y′ = xˣ(ln x + 1). The base and exponent both moving is a different animal.
"d/dx aˣ = x aˣ⁻¹." Constant base, variable exponent — the mirror error. Correct: aˣ ln a. The power rule handles xⁿ; the exponential rule handles aˣ; confusing the two is the most common derivative error in first-year work.
"(3x + 1)⁵ differentiates to 5(3x + 1)⁴." Missing the chain factor: 5(3x+1)⁴·3 = 15(3x+1)⁴. The power rule applies to the outer power; the inner function bills separately.
"The rule is proved by pattern from n = 1, 2, 3." Patterns suggest; the binomial-theorem limit (§2) proves — and the distinction is exactly what separates knowing calculus from remembering it. The extensions above show why the proof must be re-earned for each new exponent class.
7. Where this shows up
Polynomials are the local language of everything smooth, so the power rule is the engine of Taylor series: dᵏ/dxᵏ of xⁿ at 0 delivers the k! factors that define every coefficient in every expansion you will ever use. Economics runs on it (marginal cost = C′(q) for polynomial cost models); physics differentiates power-law potentials (F = −dU/dr for U ∝ r⁻¹ gives the inverse-square law in one line); numerical analysis builds finite-difference formulas by differentiating interpolating polynomials; and the power-rule calculator evaluates a·n·xⁿ⁻¹ live — the theorem of this chapter, wearing an interface.
8. Practice problems
P1. Differentiate f(x) = 7x⁵ − 4x³ + 9x − 12. Solution: 35x⁴ − 12x² + 9.
P2. Find the tangent to y = 1/√x at x = 4. Solution: y = x^{−1/2}, y′ = −½x^{−3/2}; slope −1/16; point (4, ½): y = ½ − (x−4)/16.
P3. For what x does y = x³ − 12x have horizontal tangents? Solution: y′ = 3x² − 12 = 0 → x = ±2.
P4. Differentiate y = x^{π}. Which proof from §4 licenses it? Solution: πx^{π−1}; the e^{n ln x} argument — the only one covering irrational exponents.
P5. Using the limit definition only, differentiate x⁴ at x = 1 and confirm against the rule. Solution: ((1+h)⁴−1)/h = 4 + 6h + 4h² + h³ → 4 = 4·1³ ✓.
9. Going deeper
The proof's mechanism — expand, keep the term linear in h, discard the rest — is the entire concept of the derivative as linearisation, and it scales: in several variables the "coefficient of h" becomes the gradient; for functionals it becomes the variational derivative behind Euler–Lagrange equations. Even fractional calculus honours the pattern: the Riemann–Liouville derivative of xⁿ is Γ(n+1)xⁿ^{−α}/Γ(n−α+1), collapsing to the power rule when α = 1 because Γ threads the factorials continuously. One binomial expansion in a first course; the shape of differentiation everywhere after.
10. Historical context
The rule predates "the calculus." Cavalieri, Fermat, Pascal and Wallis, computing areas under y = xⁿ in the 1630s–50s, effectively established the integral power rule (area = xⁿ⁺¹/(n+1)) case by case — Fermat's method of adequality also delivered tangents to these curves, the derivative rule in embryo, decades before Newton and Leibniz. Newton's fluxions (1660s) and Leibniz's differentials (1670s–80s) then made it a one-line consequence of general machinery; Leibniz's dxⁿ = nxⁿ⁻¹dx appears in his foundational 1684 paper. The modern proof had to wait for the limit concept itself — Cauchy's Cours d'analyse (1821) and Weierstrass's ε-δ formulation — because before rigorous limits, discarding the O(h) terms was precisely the "ghosts of departed quantities" Berkeley mocked in 1734. The rule was thus used confidently for 150 years before it could be proved honestly: a useful reminder that mathematical practice and mathematical rigour travel on different schedules.
11. Another way to see it: two more proofs
Induction via the product rule. Base case n = 1: d/dx x = 1 from the limit definition directly. Inductive step: assume (xᵏ)′ = kxᵏ⁻¹; then xᵏ⁺¹ = x·xᵏ and the product rule gives (xᵏ⁺¹)′ = 1·xᵏ + x·kxᵏ⁻¹ = (k+1)xᵏ ∎. Clean, binomial-free, and it displays the rule as a shadow of the product rule's structure — differentiation acting like a derivation on the algebra of polynomials, in the abstract-algebra sense that later mathematics makes literal.
The factorisation route. Use the algebraic identity aⁿ − bⁿ = (a − b)(aⁿ⁻¹ + aⁿ⁻²b + … + bⁿ⁻¹) with a = x + h, b = x:
No binomial theorem, no discarded remainder — the difference quotient is rewritten exactly and the limit taken term by term. Many analysts consider this the cleanest first proof; it is also the one that generalises to (f(x) = xⁿ over any field) and connects to the divided-difference identities of numerical analysis.
12. Frequently asked questions
What is d/dx x⁰? x⁰ = 1 (for x ≠ 0), a constant: derivative zero — and the rule agrees, 0·x⁻¹ = 0. No paradox, though the formula's x⁻¹ warns you x = 0 needs care.
Does the rule hold at x = 0 for fractional exponents? Sometimes: x^{3/2} has derivative (3/2)x^{1/2}, fine at 0 (one-sided); x^{1/3} has a vertical tangent at 0 — the derivative (1/3)x^{−2/3} correctly blows up. The formula is truthful even about its own failures.
Why is the integral rule "add one, divide" the mirror image? Because integration inverts differentiation: if (xⁿ⁺¹/(n+1))′ = xⁿ by the power rule, then ∫xⁿdx = xⁿ⁺¹/(n+1) + C. The lone exception n = −1 (division by zero) is exactly where the logarithm lives — the gap in the power family is ln x's job description.
Is there a power rule for other "derivatives"? Yes, systematically: finite differences give Δ(x)_n = n(x)_{n−1} for falling factorials (the discrete power rule underlying summation calculus), and fractional derivatives give the Γ-function version quoted in §9 — each calculus keeps the rule's shape and swaps the notion of power.
13. Further practice
P6. Prove d/dx x⁵ by the factorisation route explicitly. Solution: the sum has five terms, each → x⁴: derivative 5x⁴ ✓.
P7. Using induction's logic backwards, deduce the derivative of x⁻¹ from the product rule and x·x⁻¹ = 1. Solution: 0 = (1)′ = 1·x⁻¹ + x·(x⁻¹)′ → (x⁻¹)′ = −x⁻² — the negative-exponent rule earned from the identity, no limits needed.
P8. A falling-factorial exercise: with (x)_3 = x(x−1)(x−2), verify Δ(x)_3 = (x+1)x(x−1) − x(x−1)(x−2) = 3(x)_2. Solution: factor x(x−1): x(x−1)[(x+1) − (x−2)] = 3x(x−1) = 3(x)_2 ✓ — the discrete power rule, one line.
P9. Find all points where y = x⁴ − 8x² has tangent slope 0 and classify them via the second derivative. Solution: y′ = 4x³ − 16x = 4x(x²−4): x = 0, ±2; y″ = 12x² − 16: y″(0) < 0 local max (0,0); y″(±2) > 0 minima (±2, −16).
14. Worked exam problem
Problem. (a) For f(x) = 2x³ − 9x² + 12x − 3, find and classify all critical points and give the tangent line at x = 0. (b) A particle moves with s(t) = t⁴ − 4t³ (metres, seconds): find velocity and acceleration functions and all times the particle is at rest. (c) A cost model C(q) = 0.02q³ − 1.2q² + 30q: find the marginal cost at q = 50.
Solution. (a) f′ = 6x² − 18x + 12 = 6(x−1)(x−2): critical points x = 1, 2. f″ = 12x − 18: f″(1) = −6 < 0 → local max (1, f(1)) = (1, 2); f″(2) = +6 → local min (2, 1). At x = 0: slope f′(0) = 12, point (0, −3): tangent y = 12x − 3. (b) v = s′ = 4t³ − 12t² = 4t²(t − 3): rest at t = 0 and t = 3 s; a = v′ = 12t² − 24t. (c) C′(q) = 0.06q² − 2.4q + 30; C′(50) = 150 − 120 + 30 = 60 — each additional unit near q = 50 costs about 60. One rule, term by term, powering optimisation, kinematics and economics in a single sitting: this ubiquity is why the proof deserved four different demonstrations.
15. Key takeaways
The power rule d/dx xⁿ = nxⁿ⁻¹ is proved for positive integers by the binomial limit (or the cleaner factorisation identity, or induction through the product rule), extends to negative exponents by the quotient structure, to rationals by implicit differentiation, and to all reals via e^{n ln x}. Its constant companions: it needs a constant exponent (xˣ and aˣ are different animals), a bare base (chain rule otherwise), and it mirrors into the integral rule with n = −1 reserved for the logarithm. It is the term-by-term engine of all polynomial calculus and every Taylor series.
16. Where to go next
See the rule at work in the power-rule calculator and the broader power-rule calculator; the vertex problems it solves by calculus are solved by algebra alone in completing the square — comparing the two routes on the same parabola is a fifteen-minute exercise that ties this module together. Definitions of limit, derivative and tangent wait in the dictionary.